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Vladimir [108]
3 years ago
10

Suppose we have a 12.2 L sample containing 0.50 mol oxygen gas at a pressure of 1 atm and a temperature of 25 degrees celcius. I

f all this oxygen were converted to ozone at the same temperature and pressure, what would be the volume of the ozone?Balanced equation: 3O2 → 2O3a. 6.5Lb. 2.8Lc. 8.1Ld. 7.4L
Chemistry
1 answer:
snow_lady [41]3 years ago
6 0

Answer:

c. 8.1 L

Explanation:

Given that:-

Moles of oxygen gas = 0.50 mol

According to the reaction shown below as:-

3O_2\rightarrow 2O_3

3 moles of oxygen gas on reaction gives 2 moles of ozone

Also,

1 mole of oxygen gas on reaction gives 2/3 moles of ozone

So,

0.50 mole of oxygen gas on reaction gives \frac{2}{3}\times 0.50 moles of ozone

Moles of ozone = 0.3333 mol

Pressure = 1 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25.0 + 273.15) K = 298.15 K  

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K  

⇒V = 8.1 L

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Hydrocarbons consisting of short carbon chains are ________.
Tresset [83]

Answer:

A

Explanation:

Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.

As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain

4 0
3 years ago
A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is us
Zinaida [17]

Answer:

A. 96.3 mg/dL

Absolute error: 5.7 mg/dL

Relative error: 5.6%

B. 97.2 mg/dL

Absolute error: 4.8 mg/dL

Relative error: 4.7%

C. 104.8 mg/dL

Absolute error: 2.8 mg/dL

Relative error: 2.7%

D. 111.5 mg/dL

Absolute error: 9.5 mg/dL

Relative error: 9.3%

E. 110.5 mg/dL

Absolute error: 8.5 mg/dL

Relative error: 8.3%

Explanation:

The formula for the absolute error is:

Absolute error = |Actual Value - Measured Value|

The formula for the relative error is:

Relative error = |Absolute error/Actual value|

In your exercise, we have that

Actual Value = 102.0 mg/dL

A. 96.3 mg/dL:

E_{ABS} = |102.0 - 96.3| = 5.7mg/dL

E_{R} = \frac{5.7}{102} = 0.056 = 5.6%

B. 97.2 mg/dL

E_{ABS} = |102.0 - 97.2| = 4.8mg/dL

E_{R} = \frac{4.8}{102} = 0.047 = 4.7%

C. 104.8 mg/dL

E_{ABS} = |102.0 - 104.8| = 2.8mg/dL

E_{R} = \frac{2.8}{102} = 0.027 = 2.7%

D. 111.5 mg/dL

E_{ABS} = |102.0 - 111.5| = 9.5mg/dL

E_{R} = \frac{9.5}{102} = 0.093 = 9.3%

E. 110.5 mg/dL

E_{ABS} = |102.0 - 111.5| = 8.5mg/dL

E_{R} = \frac{8.5}{102} = 0.083 = 8.3%

4 0
3 years ago
Convert 682000000000 kmol to Tmol
Elena-2011 [213]

Answer:

682 teramoles

Explanation:

5 0
2 years ago
Read 2 more answers
Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? Which of these bes
Pani-rosa [81]
Nitrogen monoxide has 1 oxygen atom and
Nitrogen dioxide has 2 oxygen atoms
3 0
2 years ago
For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-
Bond [772]

Answer:

C_t=0.165M

Explanation:

From the question we are told that:

Slope K=0.056 M-1 s -1

initial Concentration C_1=2.2M

Time t=100

Generally the equation for Raw law is mathematically given by

\frac{1}{C}_t=kt+\frac{1}{C}_0

\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0

C_t=0.165M

4 0
3 years ago
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